Regex operation optimisation

perlpal has asked for the wisdom of the Perl Monks concerning the following question:

Hi Perl Monks ,

I need to perform multiple operations on a string

- remove special characters like (% ,(,),*)

- replace continuous white spaces with a single underscore

- replace all uppercase letters with lowercase letters

Currently , i have written the following code snippet to accomplish the same


                $string =~ s/ \(.*\)//g;
                $string =~ s/ {1,}/_/g;
                $string = lc ($string);
[download]

Is there a more optimized approach for the same?

Comment on Regex operation optimisation Download Code

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Re: Regex operation optimisation by jethro (Monsignor) on Jul 02, 2010 at 09:57 UTC
if you need to remove only matching (), your code won't do. Since '.' will match any character including ')' and since it is a greedy match, it will only eliminate the outmost parenthesis and (!)everything in between. Also the space before the parenthesis would have to match. You could use `s/$[^)]$//g #eliminate anything in parenthesis or s/$([^)])$/$1/g #remove matching parens but not the content` [download] but you would have to repeat that regex until no match was found anymore. Except if you only have only one level of parenthesis, then one execution of the regex would be enough removing characters can be done with one regex using a character class: `s/[%*()]//g;` [download] even parenthesis if you don't mind that they don't match you could simply use `+` instead of `{1,}` UPDATE to correct a very silly mistake indeed: changed ? to +	[reply] [d/l] [select]
Re^2: Regex operation optimisation by number2 (Initiate) on Jul 02, 2010 at 11:12 UTC
I think that `?` means `{0,1}`. Possibly `+` would be more appropriate. Also, if you wish to eliminate all whitespace characters, which includes newlines, tabs, etc., then `\s+` might be the way to go.	[reply] [d/l] [select]
Re^2: Regex operation optimisation by oldmanwillow (Novice) on Jul 02, 2010 at 12:31 UTC
A non-greedy quantifier could be used to match only the first closing parenthesis. s/$.*?$//	[reply]
Re^3: Regex operation optimisation by JavaFan (Canon) on Jul 05, 2010 at 10:40 UTC
That turns `"foo (bar (baz) qux) quux"` into `"foo qux) quux"`. I'd be very surprised if the OP wants that to happen.	[reply] [d/l] [select]
Re: Regex operation optimisation by jwkrahn (Abbot) on Jul 02, 2010 at 12:31 UTC
- remove special characters like (% ,(,),)* `$string =~ tr/%()//d;` [download] - replace continuous white spaces with a single underscore* `$string =~ tr/ /_/s;` [download]	[reply] [d/l] [select]
Re^2: Regex operation optimisation by marinersk (Priest) on Jul 02, 2010 at 21:39 UTC
Odd. I would have done - replace continuous white spaces with a single underscore `$string =~ s/\s+/\_/g;` [download] The question I have; is it simple preference or is there a technical reason you use the /s approach? I sense an opportunity for me to learn something, Obi Wan .	[reply] [d/l]
Re^3: Regex operation optimisation by jwkrahn (Abbot) on Jul 02, 2010 at 23:36 UTC
perlpal's example was `$string =~ s/ {1,}/_/g;` which matches a space character while your example `$string =~ s/\s+/\_/g;` uses the whitespace character class (`\s`) which matches `"\t"`, `"\r"`, `"\n"` and `"\f"` as well as the space character. In general, the `tr///` operator is more efficient then the `s///` operator so I try to use it when appropriate.	[reply] [d/l] [select]
Re^4: Regex operation optimisation by marinersk (Priest) on Jul 06, 2010 at 21:05 UTC